Michal, would you mind writing out the proof? In particular how do you show that sets containing zero are fixed?

How can you compute the inverse of $a_p^{k/2}$? It should be equal to itself if you quotiented by $a_p^k$, but this requires knowing that it's finite order.

About the OP: if you are comparing ZFC natural numbers to real life as in "if you take two apples and two apples you get four apples", then arithmetic consistency is not a formal property but it is clearly not the same as consistency (unless you think the theory of real numbers can't be consistent because it also doesn't model real life natural numbers). But if you are treating it as a formal statement this requires some metatheory whose definition of the naturals may or may not reflect reality too. I guess this is what Toby meant by "you have to apply this across the board."

"Both are of the form ∀n,P(n) for decidable P." Can you elaborate further? It seems to me that there is no way to disprove "there are infinitely many twin primes" by a counterexample, whereas there is for "ZFC is consistent."

Actually it does work out: note that $0^2x = (0^2 + 0^2)x = 0^2$ for all $x$. So if $0 \cdot 0 \ne 0$, take any nonzero $x$ as above. $0^2 + 0^2$ is also nonzero by hypothesis, so we can cancel $x$ to get $0^2 = 0^2 + 0^2$ so $0^2 = 0$ and we have a ring with identity.

I don't have access to links 2/3/4. Can you post something of their contents? Are there any interesting results?

Are there any known examples of vertex-transitive (and connected, depending on your definition of Cayley) graphs that have indegree = outdegree but aren't Cayley graphs? If I understand John correctly, the answer is no, although the Petersen graph cannot have indegree = outdegree because the total degree is odd.

And you can also exclude algebraic irrational $\beta$ in the same way using Gelfond-Schneider.

Most rationals don't work. Note that the above argument shows $\beta = 2\log_2(L)$ where $L$ is irrational or equal to 1. This includes $2(\mathbb{Q} \setminus \mathbb{Z}) = \mathbb{Q} \setminus 2\mathbb{Z}$: let $L = 2^q$ where $q \in \mathbb{Q} \setminus \mathbb{Z}$. But consider $x \mapsto x^{a/b}$ where $b/a \notin \mathbb{N}$. Then $4(2^{b/a})^{a/b} - 2(4^{b/a})^{a/b} = 0$ but $\frac{4^{b/a}}{2^{b/a}} = 2^{b/a} \notin \mathbb{Q}$. This leaves $\beta = 1/n$ or irrational.

This does not seem to consider the case when $f(\sqrt{2}x) = f(x)$. OP mentioned constant functions, which do satisfy the criterion because a singleton is always linearly independent. If we are considering sequences instead, then constant functions don't work.

Here are some more examples. They actually define exponential field in such a way that it implies this condition, rather than the other way around.